Question

Find the 6th term of a geometric sequence with t1=5 and r = -1/2

Answer 1

`\bf n^(th)\textit{ term of a geometric sequence}\\\\ t_n=t_1\cdot r^(n-1)\qquad \begin{cases} n=n^(th)\ term\\ t_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ n=6\\ t_1=5\\ r=-(1)/(2)\\ \end{cases}\implies t_6=5\left(-(1)/(2) \right)^(6-1)`

Answer 2

Same as before, to make the formula do

t(1) * (r)^n-1
So 5 * (-1/2)^n-1
Therefore
5*(-1/2)^5
0.078125